= Solution
The critical points of $f''$ occur where $f'''=0$, namely $-1,0,1$. Its sign on the four intervals separated by these points is, in order, positive, negative, positive, negative. Thus $f''$ has local maxima at $\pm1$ and a local minimum at zero. Their values are
$$
f''(-1)=f''(1)=\frac12,\qquad f''(0)=-2.
$$
Since $f''(x)\to0$ at both ends of the real line, these are also \b[the global maximum $1/2$ and the global minimum $-2$]. The sign analysis establishes extremality; merely solving $f'''=0$ would not.
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